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Power Factor


vluc

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Maybe one of our engineer/physics folks can answer this off-the-wall question, but is there a way to translate our Power Factor into a measure of energy such as foot-pounds or joules?

I'm guessing it may change depending on bullet weight and such so that the foot pounds for a 125 PF with a 115 grain bullet is not the same as a 135 or a 147 in 9mm. Likewise a 130 PF with a 180 grain 40 would be different from a 130 PF 9mm. Then again, maybe not.

If that is so, what then becomes the most accurate measure to calibrate steel? Should it be agreed upon that a 115 grain bullet is the minimum floor to use or do we calibrate for foot pounds?

Just got this wild-hair and thought I'd ask!

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Too much free time!

http://my.stratos.net/~thedump/odds/KEvsBM.gif

mcb

ETA but remember its the momentum of the bullet (ie PF) at impact with the steel that will determine how well the steel is knocked down not how much kinetic energy the bullet has. Two bullet of equal momentum, even if they have different mass and thus different amounts of kinetic energy, should impart the same momenum to a steel target. This assumes both bullets have similar inelastic collisions with the steel.

So in theory a 115 grain 9mm bullet at a 125PF should knock a peice of steel down the same as a 230 grain 45 bullet with a 125 PF even though the 115 grain 9mm bullet has twice the Kinetic energy.

{edit: picture to a link...as it was too wide and blew out the screen}

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Power factor is Momentum.

Momentum = Mass * Velocity

Kinetic Energy = (1/2) * Mass * Velocity * Velocity

Bullet interactions with steel plates are best described by the conservation of momentum, not kinetic energy, because heat (a form K.E.) does not act to move the plate.

Formula for conservation of momentum:

m1*v1 + .... + mn*vn = m1'*v1' + .... mn'*vn'

Personally, I wouldn't worry about unit conversions which is the only difference between Power Factor and Imperial Units. It will only change the number, not the result.

Finally, power factor is probably not the best measure of bullet effectiveness (nor is the raw K.E.), but it is the best measure we have of relative recoil. I think we just need to leave things alone and let them be simple. ;)

FWIW...

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Just off the top of my head, Energy is a function of mass x velocity.

Power factor is basically the same equation, bullet weight x velocity.

Without doing any real analysis I'd say that any two loads that have the same power factor are going to have similar muzzle energy.

The only difference is this.

A lighter / faster load may have the same energy as a heavier / slower load at the muzzle but will retain less energy down range.

High velocity loads bleed energy at a higher rate until the bullet becomes subsonic.

This however is splitting hairs.

At the ranges we shoot it should make no practical difference.

For the purposes of calibrating steel it shouldn't matter.

A 125 power factor is a 125 power factor, regardless of what bullet weight you use to attain it.

Tls

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Just off the top of my head, Energy is a function of mass x velocity.

Power factor is basically the same equation, bullet weight x velocity.

Energy is mass x velocity squared... that's a *huge* difference between the two....

Read EricW's post... he knoweth what he speaketh :D

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The original measure of power for IPSC was a ballistic pendulum using a steel bob. The impact of the bullet swung the pendulum arm and the degree of arc was noted. Minor was calibrated with a 9mm, and Major with a 45 Commander with 230gr Ball.

This was nullified when harder bullets were used. The harder bullets imparted a greater swing to the pendulum, even though the power factor, as now calculated, may not have been up to the 230gr Ball ammo. This showed that this form of ballistic pendulum measured impact much more than momentum.

As others have noted, the current power factor is a momentum figure, but with inconsistent units. The values used to calculate the momentum (mass and velocity) are the same for calculating kinetic energy.

Guy

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So in theory a 115 grain 9mm bullet at a 125PF should knock a peice of steel down the same as a 230 grain 45 bullet with a 125 PF even though the 115 grain 9mm bullet has twice the Kinetic energy.

In theory, yes, in reality no. Because the interaction really is an impulse, and the 45 bullet delivers a greater impulse to the plate than the 9mm. You have to compare impulse curves in order to understand the imparted changes in momentum.

If people are really interested in the math, most of this is covered in the first two intro college physics courses and in Statics and Dynamics (usually two separate courses) in the mechanical engineering curriculum. In addition to forces, vectors, momentum, and impulses, you'll get to do neat stuff like basic orbital mechanics using conservation of momentum.

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Guy pointed out the factor that makes a large difference in actual energy imparted versus calculated energy imparted to any object a projectile collides with.

Assuming the angle of incidence is the same and the momentum is the same for each compared projectile, the one that disintegrates easiest will impart the least energy to the plate because more of it's energy is converted to/dissipated as heat.

You would have to take projectile construction into account to get the correct calculated result as compared to what a ballistic pendulum will tell you when struck with differing projectiles. The velocity difference alone will disintegrate similarly built but different weight projectiles at differing rates anyway. An index of frangibility at a specific velocity for each projectile needs to be applied to the calculation if we wish to compare the actual momentum transfer capability of different projectiles at different velocities.

Bullets hitting steel plates are not 100% labratory grade elastic collisions so a formula assuming this is skewed to start with ;-)

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But a bullet impact is close enough to a truly inelastic collision that I dare say you could not measure the difference in momentum imparted to the steel between the 115 grain 9mm and 230 grain 45 if both where loaded to a 125 PF. In both cases the bullet completely splats (very technical term) on the steel. Thus all of the momentum of the bullet is transferred to the steel minus the the component of the momentum particles have that is perpendicular to the plate. Since with good steel and normal copper lead constructed bullets all the fragment come off the plate nearly parallel to the face of the steel all the momentum has to be transferred to the steel. Within the confines of copper and lead bullet construction there should be very little difference between the 115gr bullet going splat and the 230 grain bullet going splat. Assuming the instant before impact both bullet have the same momentum than just after the impact the steel will also have nearly the same momentum there is just not that many other place for the momentum to go. Kinetic energy never enters into these equation if you get into the math.

mcb

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In both cases the bullet completely splats (very technical term) on the steel. Thus all of the momentum of the bullet is transferred to the steel minus the the component of the momentum particles have that is perpendicular to the plate.

Just an observation, but I've seen many a .45 bound off a plate or popper, but not many 38 super's...

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In both cases the bullet completely splats (very technical term) on the steel. Thus all of the momentum of the bullet is transferred to the steel minus the the component of the momentum particles have that is perpendicular to the plate.

Just an observation, but I've seen many a .45 bound off a plate or popper, but not many 38 super's...

Now in that case you would get a greater momentum transfer to the plate. The fact that you impart a momentum vector to the 45 bullet in the opposite direction it impacted the plate would mean a larger amount of momentum was imparted to the plate. Although I suspect that the velocity is rather low so the extra momentum would not be great but measurable.

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After consulting with my (really old :( ) college textbooks .....

I think EricW's first response best answers VLUC's original question.

Conservation of momentum (not energy) most accurately describes the interaction between a projectile and a steel target. Since PF is in essence a measure of momentum (Mass x Velocity), that would be the best way to define steel calibration.

At a constant PF, differing bullet weights has a minimal effect - similar to or less than the effects of hundreds of other variables like bullet material and design, popper condition, ambient conditions, etc.

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Okay,

You guys forced me into this.

The original question was:

In the context of calibrating a popper, is there a difference in using a 147 gr bullet at 125 PF vs doing the calibration with a 115 gr bullet at 125 PF.

I could dust off my physics book and do an in depth engineering analysis to see just what the actual differences are.

However, after taking into account, moment of inertia, modulus of elasticity, and rockwell hardness of a cantilevered simple beam aka (the popper), momentum, mass, velocity, jacket hardness, bullet alloy, ballistic coefficent of the bullet, angles of incidence and refraction, coefficient of friction on the hinge of the popper, rotational torque due to rate of twist, atmospheric pressure, rotation of the earth, gravitional effects due to alignment of the stars, sun and moon, ......

The rocket science will still show no Practical difference.

:wacko::P:D

Who gets the bill?

Tls

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I'm convinced!

Now,

will someone please load up some 115 grain 9mm and some 230 grain .45's using the same type bullets to 125 pf and shoot a popper to see if there is a difference?

I just love it when the experimental physicists prove the theoretical physicists correct. :)

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Do the test with a 38 super and a 45 both loaded to 170-175 PF. Use a video camera and a background grid to record the angular velocity of the popper as it falls.

I've got money on which one drives a popper down faster. ;)

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The fair test of this postulate would be to load some FMJ 9mm projectiles up to 220 PF and compare them against similarly constructed FMJ's in .45 at 220 pf.

125 is way to low an energy level to bring out the kinetic energy effect on the 9x as the velocity is still mundane.

An even better real world test would be a 55gr .223 at 175 pf versus a 220 grain .45 at 175 pf. Both at 10-15 yards on a fresh steel plate surface mounted to a ballistic pendulum (with operator safety shielding in place of course) to prevent angle of incidence differences from skewing the results. This is enough of a difference in type of energy present to show some real measurable result.

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Which brings us back to my idea of a calibration hammer.

A weight (the hammer) swung on a fixed length rigid pendulum will impart the same force day after day to EXACTLY the same place on the face of the steel popper.

All I need is a length of moment arm and the weight of the hammer and I will make up a sample.

Figure that it will have two elevation positions and a small leveling bubble like a camera tri-pod. Hold the unit at the front of the popper, retract the weight to the horizontal position and let it go. perfect 125 PF every time! And you can calibrate steel without clearing the range, waiting for the mandated start time where there are noise ordinances, or calling for and RM.

Jim

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There is no need to go berzerk with the PF. 175 will demonstrate everything needed.

Hitting a popper in same area +/- an inch or two at seven yards is non-challenge. If I had a super, I'd be headed to the range already just end the discussion. Anybody that's shot steel for any length of time has watched whole 45 slugs slide off steel plates and 38's vaporize into shrapnel. The problem just isn't that complicated.

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